Given the parametric equations, x = 2t + 5 and y = 3t^2 - 1, find the point on the graph at time, t = 4.
1 answer:
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Answer:
C
Step-by-step explanation:
It usually works best to use the polynomial with fewer terms as the multiplier. A row of partial products is written for each term of the multiplier, so the fewer terms will result in fewer rows of partial products.
In order to keep like terms together, it is preferable to allocate a separate column of the multiplication tableau to each power of the operands or product. This means we want to make note of the fact that the cubic multiplicand has a coefficient of 0 for its x^2 term.
The best setup is the one shown in the attachment.
Answer:
y=2x
Step-by-step explanation:
y = kx is the format of the graph we are looking for (as told in the instructions) where k is the variable we are supposed to find (like y = x, y=2x, y=3x, etc)
Note that as x increases by 1, y increases by 2. We can see that by writing a table of the points of points that the line given intersects (picture below). We can see that by the values of the x-axis and the y-axis. Why does this happen? This is because y =2x for each value of x. We notice that in our table.
Therefore, the equation is y=2x. I hope this helps! If you still have questions, ask me in the comments!
